Question

Light with an energy flux of $20 \,W/cm^2$ falls on a non-reflecting surface at normal incidence. If the surface has an area of $30 \,cm^2$, the total momentum delivered (for complete absorption) during $30$ minutes is

• A. $36 \times 10^{-5}\,kg\,m/s$
• B. $36 \times 10^{-4}\,kg\,m/s$
• C. $108 \times 10^{4}\,kg\,m/s$
• D. $1.08 \times 10^{7}\,kg\,m/s$

Answer 1

Answer: (B)

For complete absorption,
$p = \frac{U}{c}$
where $U =$ Energy carried by light $= IAt$
$\therefore \quad p = \frac{IAt}{c}$
Here, $I = 20\, W/cm^{2}$, $A = 30\, cm^2$, $t = 30\, min = 1800 \,s$
So, $p = \frac{20 \times 30 \times 1800}{3 \times 10^{8}}$
$= 36 \times 10^{-4}\,kg\,m\,s^{-1}$